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If $\alpha=\large\frac{5}{2 ! 3}+\frac{5.7}{3 ! 3^2}+ \frac{5.7.9}{4 ! 3^3}+........,$ then $ \alpha^2+4 \alpha=$

\[\begin {array} {1 1} (1)\;21 & \quad (2)\;23 \\ (3)\;25 & \quad (4)\;27 \end {array}\]

1 Answer

(2) 23
answered Nov 7, 2013 by pady_1
Can u plz explain the steps
 
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